AQuaternionFrameworkforRobustColorImages Completionmng/mng_files/QMC.pdfperforming image completion separately in each color channel. Keywords: Color images, quaternion, matrix recovery,

A Quaternion Framework for Robust Color Images

Completion

Zhigang Jia∗ Michael K. Ng† Guang-Jing Song‡

May 1, 2018

Abstract

In this paper, we study robust color images completion and provide a rigorous anal-

ysis for provable estimation of color images from a random subset of their corrupted

pixels. Our analysis is focused on the use of quaternion framework as a representation

of color images. We solve a convex optimization problem, which minimizes a nuclear

norm of quaternion matrix which is a convex surrogate for a quaternion matrix rank,

and the l1-norm of sparse quaternion matrix entries. We show that under incoher-

ence conditions that a quaternion matrix can be recovered exactly with overwhelming

probability, provided that its rank is sufficiently small and the corrupted entries are

sparsely located. Numerical examples are presented to illustrate our theoretical results.

The results of noisy color images completion are also given to show the performance

of the proposed approach is better than the testing methods using the unfolding and

performing image completion separately in each color channel.

Keywords: Color images, quaternion, matrix recovery, low rank, convex optimization

1 Introduction

In the last decade, there is a lot of research interest and applications in science and engi-neering for robust recovery of low-rank matrices from incomplete and/or corrupted entries,see for instance [1, 3, 4, 5, 30, 36]. Among these low-rank recovery problem, there aremany problems involved in image processing. Usually, the trick is to stack all the imagepixels as column vectors of a matrix, and recovery theory and algorithm are applied to theresulting matrix which is a low-rank or an approximate low-rank matrix. Theoretically, ithas been shown that under some mild assumptions, efficient techniques based on convexprogramming to minimize the nuclear norm, as an approximation for the matrix rank, canaccurately recover low-rank matrices [4, 5, 25, 28, 36].

Nowadays, color images appear commonly in many image processing tasks. In this paper,we are interested in robust color images completion. This paper considers a problem oflearning a low-rank color image from undersampled and/or corrupted measurements. Such

∗School of Mathematics and Statistics, Jiangsu Normal University, Xuzhou 221116, P.R. China. Email:[email protected]

†Corresponding author. Department of Mathematics, Hong Kong Baptist University, Hong Kong. E-mail:[email protected]

‡School of Mathematics and Information Sciences, Weifang University, Weifang 261061, P.R. China.Email: [email protected]

1

color image processing problem arises in a wide range of imaging applications in which colorimage pixels contain missing values and/or gross errors due to many reasons. The reasonsinclude transmission loss, sensor failure and hardware/software malfunction.

A color image contains red, blue and green channels. One approach is to conduct robustmatrix completion for each channel, and then the final processed image is obtained by com-bining the results from red, green and blue channels. The disadvantage of this approach isthat the inter-channel information is not considered and used for robust color image com-pletion procedure. On the other hand, a color image with red, blue and green channels canbe naturally regarded as a third-order tensor. Each frontal slice of this third-order tensorcorresponds to a channel of the color image. In general, we can apply tensor-based com-pletion [9, 20, 31] and robust tensor principal component analysis [12, 19] to deal with thisspecific color image tensor. However, the recovery theory for low-rank tensor completionproblems is not well-established compared with low-rank matrix completion problems. Themain reason is that tensor rank has different definitions in the literature. The CANDE-COMP/PARAFAC (CP) decomposition [6, 15] approximates a tensor as sum of rank-oneouter products and the minimal number of such decomposition is defined as the CP-rank.However, computing the CP-rank of a specific tensor is NP-hard in general [16]. Tuckerdecompositions [32] reveal the algebraic structure in the tensor data with the notion of rankextended to multilinear rank, which is expressed as a vector of ranks of matrix factors. It isclear that Tucker decompositions cannot offer the best low-rank approximation to a tensor.

In this paper, we make use of quaternion to represent color images and study the prob-lem of robust color images completion under the quaternion framework. The quaternionapproach is to encode the red, green and blue channel pixel values on the three imaginaryparts of a quaternion. The use of quaternion for color images representation has been stud-ied in the literature, see [7, 27, 29]. The main advantage is that color images can be studiedand processed holistically as a vector field, see [7, 27, 29]. Many gray-scale image processingmethods have been extended to color image processing using the quaternion algebra, forinstance Fourier transform [27], wavelet transform [8], principal component analysis [38],independent component analysis [34] and etc. A color image is represented as a quaternionmatrix and the singular value decomposition of a quaternion matrix exists [24]. Similar toreal/complex matrix case, the Eckart-Young-Mirsky low-rank approximation theorem [11]can be applied to the case of quaternion matrices [37]. Because of such excellent property,it is expected to be useful to study and analyze robust color images completion under thequaternion framework. We show that one can obtain an exact recovery of a quaternion ma-trix with high probability by simply solving a convex program where the objective functionis a weighted combination of nuclear norm, serving as a convex surrogate for quaternionmatrix rank, and the ℓ1-norm of sparse quaternion matrix entries. The conditions underwhich our result holds, similar to the classical real matrix incoherence conditions [4, 3, 18].We remark that we deal with the recovery problem for a set of quaternion numbers, sometechnical proofs are required to be established and shown in the paper.

The rest of this paper is organized as follows. In Section 2, we begin with a brief review ofquaternion matrix. The notations and some preliminaries of quaternion are also introduced.In Section 3, we give our main theoretical results and provide the idea of the proof of ourquaternion matrix recovery theorem. The detailed proofs will be presented in Appendix.In Section 4, we report numerical results to demonstrate our theoretical results. Finally,concluding remarks are given in Section 5.

2

2 Preliminaries of Quaternion

Quaternion was introduced by Hamilton [14]. Quaternion has one real part and three imag-inary parts given by

q = qr + qii+ qjj+ qkk

where qr, qi, qj , qk ∈ R and i, j and k are three imaginary units satisfying

i2 = j2 = k2 = −1, ij = −ji = k, jk = −kj = i, ik = −ki = j.

i, j and k are the fundamental quaternion units. Here a symbol of boldface indicates that itis a quaternion number, vector or matrix. When qr = 0, q is called a pure quaternion. Forsimplicity, we denote Q by a set of quaternion numbers.

Let p,q ∈ Q. The sum p+ q of p and q is

(pr + qr) + (pi + qi)i+ (pj + qj)j+ (pk + qk)k

and their multiplication pq is given by

(prqr − piqi − pjqj − pkqk) + (prqi + piqr + pjqk − pkqj)i+

(prqj + pjqr + pkqi − piqk)j+ (prqk + pkqr + piqj − pjqi)k.

The conjugate and modulus of q are respectively defined by

q∗ = qr − qii− qjj− qkk and |q| =√

q2r + q2i + q2j + q2k.

A color image with the spatial resolution of n1 × n2 pixels is represented by an n1 × n2

quaternion matrix A in Qn1×n2 as follows:

As,t = Rs,ti+Gs,tj+Bs,tk, 1 ≤ s ≤ n1, 1 ≤ t ≤ n2,

where Ri,j , Gi,j and Bi,j are the red, green blue pixel values respectively at the location(i, j) in the image.

Note that quaternion matrix-matrix multiplications can be defined similar to classicalmatrix-matrix multiplication except the multiplication between two quaternion numbers areemployed. The identity quaternion matrix I is the same as the classical identity matrix. Theinverse B of a square quaternion matrix A exists if AB = BA = I. A square quaternionmatrix is unitary if A∗A = AA∗ = I, where A∗ is the conjugate transpose of A. Thefollowings are some definitions and results for quaternion matrices. For detailed information,we refer to [37].

Definition 2.1. The maximum number of right linearly independent columns of a quater-nion matrix A ∈ Qn1×n2 is called the rank of A.

It is clear that the rank of A can refer to the maximum number of left linearly indepen-dent rows of A. Also for any two invertible quaternion matrices P and Q of appropriatesizes, A and PAQ have the same rank.

Theorem 2.1. Let A ∈ Qn1×n2 be of rank r. Then there exist two unitary quaternionmatrices U = [u1,u2, · · · ,un1 ] ∈ Qn1×n1 and V = [v1,v2, · · ·vn2 ] ∈ Qn2×n2 such that

A = UΣV∗, (1)

where Σ = diag(σ1, · · · , σr, 0, · · · , 0) ∈ Rn1×n2 , and σs (s = 1, · · · , r) are positive singularvalues of A.

3

Similar to real/complex matrix case, the Eckart-Young-Mirsky low-rank approximationtheorem [11] can be applied to the case of quaternion matrices. Because of such excellentproperty, it is expected to be useful to study and analyze robust color images completionusing quaternion.

The inner product between two quaternion matrices A and B is given by 〈A,B〉 =Tr(A∗B). The norms of quaternion vector and matrix can also be defined. For detailedresults related to the quaternion Hilbert spaces, we refer to [10] and [23].

Definition 2.2. 1. Let x = [xi] ∈ Qn be a quaternion vector. ‖x‖1 :=∑

i

|xi|; ‖x‖2 :=√

∑

i

|xi|2; and ‖x‖∞ := maxi

|xi|.

2. Let A = [ai,j ] ∈ Qn1×n2 be a quaternion matrix. ‖A‖1 := maxj

n1∑

i=1

|ai,j |; ‖A‖2 :=

max σ (A), where σ (A) is the set of singular values of A; ‖A‖∞ := maxi,j

|ai,j |; ‖A‖F =√

∑

i,j

|ai,j |2 :=√

tr (A∗A); and ||A||∗ :=r∑

s=1

σs.

The nuclear norm || · ||∗ serves as a convex surrogate for quaternion matrix rank, seeAppendix Section 6.1. We can use it to recover low-rank quaternion matrices exactly.

3 Robust Quaternion Matrix Completion

In this section, we aim to recover an n1-by-n2 quaternion matrix L0 based on that a fractionof its entries are given and these entries may be corrupted. Mathematically, we consider theentries of L0 + S0 which are in the subset Ω, can be written as

X = PΩ (L0 + S0) .

Here PΩ as the orthogonal projection onto the linear space of matrices supported on Ω ⊆[1 : n1]× [1 : n2],

PΩ(X) =

xi,j , (i, j) ∈ Ω0, (i, j) /∈ Ω.

The recovery model is given as follows: Let X be an n1-by-n2 quaternion matrix. Theminimization problem is

minL∈Qn1×n2 ,S∈Qn1×n2

‖L‖∗ + λ ‖S‖1 subject to PΩ (L+ S) = X. (2)

The above model can be applied to recover a color image from given observed and corruptedcolor image pixels. The problem is similar to the real matrix completion problem stated in[3]. In order to make the recovery possible, we consider the following incoherence conditionson the underlying low-rank quaternion matrix.

Definition 3.1. Let U be a subspace of Qn of dimension r and PU be the orthogonalprojection onto U. Then the coherence of U is defined as

µ (U) =n

rmax1≤i≤n

‖PUei‖22 ,

where ei is the i-th unit vector in Rn.

4

Let the singular value decomposition of low-rank quaternion matrix L0 in (1). Theright column and left row spaces are denoted by U and V, respectively. The incoherenceassumptions required are given as follows:

maxi

‖U∗ei‖ ≤ µr

n1, max

i‖V∗ei‖ ≤ µr

n2(3)

and

‖UV∗‖∞ ≤√

µr

n1n2, (4)

where µ is a positive constant.For simplicity, we denote n(1) = maxn1, n2 and n(2) = minn1, n2 in the following

discussion. Under the above assumption, we prove the following main theorem in the paper.

Theorem 3.1. Suppose that L0 ∈ Qn1×n2 obeys the conditions (3)-(4), Ω is uniformlydistributed among all sets of cardinality m with m = ρn1n2, each observed entry is corruptedwith probability γ independently of other entries. Then there exists a numerical constant csuch that with probability at least 1− cn−10

(1) , the solution L of (2) with λ = 1√ρn(1)

is exact,

i.e. L = L0, provided that

rank(L0) ≤ρrn(2)

µ(

log n(1)

)2 , and γ ≤ γs.

Here ρr and γs are positive numerical constants.

The proof of this theorem can be found in the next section. According to Theorem 3.1,we can recover a target quaternion matrix with probability near to one from a subset of itsentries if they are arbitrarily corrupted when the singular vectors U and V are reasonably

spread, the upper bound of the rank of quaternion matrix L0 is ρrn(2)µ−1(

logn(1)

)−2, and

the corruption term is sufficiently sparse.

3.1 The Proof

In this section, we provide of the proof of Theorem 3.1. Let us first review several conceptsand procedures required in the proof. All the proofs of the following lemmas can be foundin Appendix.

3.2 The Sampling Scheme

To facilitate our proof, we consider the Bernoulli model as a proxy for uniform sampling, theprobability of failure under Bernoulli sampling with ρ = m

n1n2is the probability of failure

under the uniform sampling. In Theorem 3.1, we consider Ω sampled according to Bernoullimodel where δi,j1≤i≤n1,1≤j≤n2 is a sequence of independent identically distributed (0 or1) Bernoulli random variables with

Prob(δi,j = 1) = ρ :=m

n1n2, (5)

where E|Ω| = m and Ω = (i, j) : δi,j = 1. Here E is the expectation operator.In (2), the available data of the form X = PΩ(L0 + S0) = PΩL0 + S′

0. There are othertwo index sets of interest. The first index set is Γ ⊆ Ω referring to the locations wheredata are available and clean, i.e., PΓX = PΓL0. Here PΓ as the orthogonal projection onto

5

the linear space of matrices supported on Γ ⊆ [1 : n1] × [1 : n2], The second index set isΩ′ = Ω/Γ referring to the locations where observed data are unreliable. The locations of Ω,Γ and Ω′ are characterized by the Bernoulli probability distributions with parameters ρ, γρand (1− γ)ρ, where 0 < γ < 1.

It is interesting to note that an quaternion random operator is a special kind of thenoncommutative random operator. The following lemma generalizes the main result of [26]from the real field to the quaternion skew field, see Appendix Section 6.2.

Lemma 3.2. Let x be a random vector in Qn, satisfying E (xx∗) = I. Let N be a positivenumber and let x1, · · · ,xN be independent copies of x. Then there exists a positive numberc1 such that

E

∥

∥

∥

∥

∥

1

N

N∑

i=1

xix∗i − I

∥

∥

∥

∥

∥

≤ c1

√

logn

N

(

E ‖x‖logN)

1log N

,

provided that the above right hand side expression is smaller than 1.

3.3 Subgradients of ℓ1 Norm and Nuclear Norm of Quaternion Ma-

trix

Let A ∈ Qn1×n2 , the subgradient of ‖A‖ can be given by

∂ ‖A‖ =

G ∈ Qn1×n2 : ‖B‖ ≥ ‖A‖+Re 〈G,B−A〉 , for all B ∈ Qn1×n2

.

Here Re(A) denote the real part of a quaternion matrix A.

Lemma 3.3. Suppose that A ∈ Qn1×n2 , the subgradient of ‖·‖1 at A supported on Ω isgiven by

∂ ‖A‖1 =

G ∈ Qn1×n2 : G = direc(A) + F,PΩ (F) = 0, ‖F‖∞ ≤ 1

, (6)

where direc(A) is an n1-by-n2 matrix with its entries given by(

ai,j

|ai,j|

)

n1×n2

.

Lemma 3.4. Suppose that A ∈ Qn1×n2 has the singular value decomposition as in (1).Then G ∈ Qn1×n2 is a subgradient of the nuclear norm at A if

G =∑

1≤k≤r

ukv∗k + F, (7)

where F have the following two properties: (a) the right column space of F is orthogonal toU and the left row space of F is orthogonal to V; and (b) ‖F‖ ≤ 1.

The proofs of the above two lemmas can be found in Appendix Sections 6.3 and 6.4respectively.

3.4 Orthogonal Projections

For a given matrix A ∈ Qn1×n2 with rank r, there exist two unitary matrices U =[u1,u2, · · · ,ur] and V = [v1,v2, · · ·vr] obtained from the singular value decompositionof A in (1). Denote Qn1×n2 = T⊕ T⊥, where T is a linear space expressed as follows:

T = UY∗ + ZV∗|Y ∈ Qn2×r,Z ∈ Qn1×r. (8)

6

Then the orthogonal projection PT onto T is given by

PT (Z) = PUZ+ ZPV − PUZPV,

where PU and PV are the orthogonal projection onto U and V, respectively. It is clear thatPT is a linear operator mapping from a matrix to an another matrix. Similarly, we definethe projection onto T⊥, the complement orthogonal space of T, by

PT⊥(Z) = Z− PT(Z).

In the proof of Theorem 3.1, we need to characterize the equivalence between two quaternionmatrices.

Lemma 3.5. Suppose that A,B ∈ Qn1×n2 , and T is defined as (8). Then 〈A,PT (B)〉 =〈PT (A) ,B〉.

The proof of Lemma 3.5 can be found in Appendix Section 6.5.In the sequel, we show with high probability that the following inequalities of random

quaternion operators hold.

Lemma 3.6. Suppose that Ω is sampled from the Bernoulli model with parameter ρ, andthe coherence obeys maxµ (U) , µ (V) ≤ µ0. Then there is a numerical constant c suchthat

ρ−1 ‖PTPΩPT − ρPT‖ ≤ c

√

µ0n(1)r logn(1)

m(9)

with probability at least 1− 3n−β

(1) where β > 1, provided that c

√

µ0n(1)r logn(1)

m< 1.

Lemma 3.7. Suppose that Ω is sampled from the Bernoulli model with parameter ρ, andthe coherence obeys maxµ (U) , µ (V) ≤ µ0. Then with high probability,

∥

∥PT − ρ−1PTPΩPT

∥

∥ ≤ ε, (10)

provided that ρ ≥ c2µ0r logn(1)

ε2n(2), where c is a constant as in Lemma 3.6.

Lemma 3.8. Suppose that Ω is sampled from the Bernoulli model with parameter ρ. Then

with high probability, ‖PΩPT‖2 ≤ ρ + ε, provided that 1 − ρ ≥ c2µ0r logn(1)

ε2n(2), where c is a

constant as in Lemma 3.6 and ε is defined as in Lemma 3.7. In particular, ‖PΩPT‖ ≤ 12

holds with high probability if ρ+ ε ≤ 14 .

Lemma 3.9. Suppose that Ω is sampled from the Bernoulli model with parameter ρ. ForZ ∈ T, then with high probability,

∥

∥Z− ρ−1PTPΩZ∥

∥

F≤ ε ‖Z‖F ,

where ε is defined as in Lemma 3.7.

Lemma 3.10. Suppose that Ω is sampled from the Bernoulli model with parameter ρ. ForZ ∈ T, then with high probability,

∥


∥

∞ ≤ ε ‖Z‖∞ , (11)

provided that ρ ≥ c2µ0r logn(1)

ε2n(2), where c and ε are defined as in Lemma 3.6 and Lemma 3.7,

respectively.

The proofs of the above five lemmas can be found in Appendix Section 6.6.

7

3.5 The Key Steps

There are several steps in the proof of Theorem 3.1. The first step is to show an optimalsolution of (2) by using orthogonal projections. The subgradients of ℓ1 norm and nuclearnorm of quaternion matrix will be considered and used. The following theorem provides theconditions for such optimal solution.

Theorem 3.11. Assume that for any quaternion matrix M,

‖PTPΓ⊥M‖F ≤ n(1) ‖PT⊥PΓ⊥M‖F

and take λ ≥ 4n(1)

. Then (L0,S′0) is the unique solution of (2) if there is a pair (W,F)

satisfyingUV∗ +W + PTD = λ(direc(S′

0) + F),

with PTW = 0, ‖W‖ < 12 , PΓ⊥F = 0 and ‖F‖∞ < 1

2 and ‖PTD‖F ≤ n−2(1).

Here PΓ⊥ as the orthogonal projection onto the linear space of matrices supported onΓ⊥ ⊆ [n1]×[n2]. The proof of Theorem 3.11 can be found in Appendix Section 6.7. The nextstep is to check the required assumptions in the theorem. By using the results in Lemmas3.6-3.10, we show tht these assumptions are valid with high probability, see Proposition 6.5in Appendix Section 6.7.

Next we show our main results in Theorem 3.1, it is sufficient to construct a pair(YL,WS) obeying

(a) PΓ⊥YL = 0, (b)∥

∥PT⊥YL∥

∥ <1

4, (c)

∥

∥PTYL −UV∗∥

∥

F≤ n−2

(1),

(d)∥

∥PΓYL∥

∥

∞ <λ

4, (e) PTW

S = 0, (f) PΩ′WS = λdirc(S′0),

(g) PΩ⊥WS = 0, (h)∥

∥WS∥

∥ ≤ 1

4, and (i)

∥

∥PΓWS∥

∥

∞ ≤ λ

4

with high probability, see [4]. Once such a pair of matrices (YL,WS) is constructed (seeAppendix Section 6.8), we can prove that there exist two matrices W and F satisfying

YL +WS = λ (direc(S′0) + F) = UV∗ +W + PTD,

where D is defined as in Theorem 3.11. The construction of YL and WS as well as theproof can be found in Appendix Section 6.8. Hence, the results of Theorem 3.1 follow.

4 Numerical Examples

In this section, we present numerical results on synthetic data and color images to validateour main theorem. All these experiments were performed in Matlab on a Mac Pro with 2.4GHz Intel Core i7 processor and 8 GB 1600MHz DDR3 memory. Here we solve the convexoptimization problem (2) by using alternating direction method of multipliers, see [4].

8

Table 1: The comparison of different recovery methods for 5% gross errors and 95% observedentries.

n rank(L0) rank(L) ‖L− L0‖F /‖L0‖FQMC RMC1 RMC2 QMC RMC1 RMC2

2 2 47 50 2.7478e-07 4.5070e-03 9.0428e-044 4 50 50 4.3244e-07 6.3223e-02 3.4267e-02

50 6 6 50 50 5.2683e-07 1.3913e-01 1.2127e-018 8 50 50 6.4845e-07 1.7792e-01 1.8226e-0110 10 50 50 9.1343e-07 2.0549e-01 2.2281e-012 2 80 93 2.0858e-07 7.6474e-07 1.3769e-066 6 100 100 3.6925e-07 8.6860e-03 4.7882e-03

100 10 10 100 100 5.6842e-07 9.2288e-02 8.7522e-0214 14 100 100 7.1509e-07 1.5172e-01 1.5889e-0120 20 100 100 8.8028e-07 2.0072e-01 2.1723e-01

4.1 Synthetic Data

We first validate the exact recovery phenomenon in Theorem 3.1 on randomly generatedquaternion matrices. We randomly generate a rank-r n-by-n quaternion matrix L0 as aproduct L0 = UrΣrV

∗r where Ur and Vr are n× r matrices, U∗

rUr = Ir , V∗rVr = Ir, and

Σr is diagonal r× r matrix with positive diagonal elements. E = E0 +E1i+E2j+E3k is arandom noise quaternion matrix, where Et (t = 0, 1, 2, 3) are independently sampled from astandard uniform distribution. Sparse noise matrix S0 is generated by randomly choosing asupport set Ω′ of size ℓ uniformly at random, and setting S0 = E on Ω′. The sparsity of S0

is denoted by γ = ℓ/n2. Let Ω be the set of observed entries. ρ = |Ω|/n2 is the percentageof observed entries.

We compare our proposed quaternion matrix completion method (QMC) with the othertwo real matrix completion methods. The first comparison (RMC1) is to apply the realmatrix completion method to the unfolding matrix of size n-by-3n [X1, X2, X3] and thesecond comparison (RMC2) is to apply the real matrix completion method to each matrixX1, X2 andX3, and then combine the results together. In QMC, we set the value of λ = 1√

ρn

given in Theorem 3.1. In RMC1 and RMC2, we set their corresponding values according totheir exact recovery results in [3], i.e., λ = 1√

3ρnand λ = 1√

ρnrespectively. The stopping

criteria of the alternating direction method of multipliers are that ‖X− L− S‖F/‖X‖F ≤10−6 and the maximum number of iterations is 5000. Table 1 reports the results withγ = 0.05 and ρ = 0.95. We see from the table that QMC performs very well. Low-rankmatrices can be recovered exactly and the corresponding relative errors are very small.However, both RMC1 and RMC2 cannot recover low-rank matrices. Indeed, their recoveredranks are significantly larger than the actual ranks.

To further validate our theoretical results, we check the recovery ability of our algorithmas a function of rank(L0), fractions of gross corruptions γ and proportion of observed entriesρ. The data are generated randomly as the above mentioned. We fix the matrix size to be50. We set ρ to be different specified values, and vary rank(L0)/n and γ to empiricallyinvestigate the probability of recovery success. For each pair (rank(L0)/n, γ), we simulate10 test instances and declare a trial to be successful if the recovered quaternion matrix L

satisfies ‖L − L0‖F /‖L0‖F ≤ 10−5. Figure 1 and Figure 2 report the fraction of perfectrecovery for each pair (black = 0% and white = 100%). We see that the smaller the

9

n=50, ρ=0.95

0.1 0.2 0.3rank(L

0)/n

0.7

0.6

0.5

0.4

0.3

0.2

0.1

γ

n=50, ρ=0.90

0.1 0.2 0.3rank(L

0)/n

0.7

0.6

0.5

0.4

0.3

0.2

0.1

γ

n=50, ρ=0.80

0.1 0.2 0.3rank(L

0)/n

0.7

0.6

0.5

0.4

0.3

0.2

0.1

γ

Figure 1: Exact recovery for varying rank rank(L0)/n (x-axis) and gross corruptions γ(y-axis) under different proportions of missing entries.

n=50, γ=0.05

0.1 0.2 0.3rank(L

0)/n

0.3

0.4

0.5

0.6

0.7

0.8

0.9

ρ

n=50, γ=0.10

0.1 0.2 0.3rank(L

0)/n

0.3

0.4

0.5

0.6

0.7

0.8

0.9

ρ

n=50, ρ=0.20

0.1 0.2 0.3rank(L

0)/n

0.3

0.4

0.5

0.6

0.7

0.8

0.9γ

Figure 2: Exact recovery for varying rank rank(L0)/n (x-axis) and missing entries ρ (y-axis).

percentage of missing values/the smaller the level of noise is, the larger the region of correctrecovery is.

Next we compare QMC, RMC1 and RMC2 by assuming a Bernoulli model for the supportof the sparse term S0 where each entry of S0 is corrupted with probability γ. The observedentries are denoted by Ω, and each position appears with probability ρ. For each (γ, ρ) pair,we generate ten random problems, and declare a trial to be successful if the recovered L

satisfies ‖L−L0‖F /‖L0‖F ≤ 10−5. Figures 3 plots the fraction of correct recoveries for eachpair (γ, ρ). The size of the matrix is 50-by-50, i.e., n = 50. Similarly, we plot different valuesof (rank(L0)/n, γ) and (rank(L0)/n, ρ) in Figures 4 and 5, respectively. It is clear that theperformance of the proposed quaternion method is better than the other two methods.

4.2 Color Images

In this experiment, we present color images completion results. The original four colorimages are shown in Figures 6, 8, 10 and 121. A standard uniform noise is independentlyand randomly added into ℓ pixel locations of red, green and blue channels of color images.The sparsity of noise components is denoted by γ = ℓ/(n1n2) where the size of image is n1-by-n2. Let Ω be the set of observed entries which are generated randomly. ρ = |Ω|/(n1n2)

1 NGC6543a image of size 600 × 600 comes from www.windows2universe.org/the universe/images/NGC6543a image.html. It is taken by the Hubble Space Telescope. Pepper image of size 512 × 512 can beused from Matlab. BSD124084 image of size 321× 481 is from Berkeley Segmentation Database (BSD) [22].Flower image of size 362 × 500 can be used from Matlab.

10

0.1 0.2 0.3 0.4γ

0.7

0.8

0.9

ρ

n=50, rank(L0)/n=0.1

(a) RMC1

0.1 0.2 0.3 0.4γ

0.7

0.8

0.9

ρ

n=50, rank(L0)/n=0.1

(b) RMC2

0.1 0.2 0.3 0.4γ

0.7

0.8

0.9

ρ

n=50, rank(L0)/n=0.1

(c) QMC

0.1 0.2 0.3 0.4γ

0.7

0.8

0.9

ρ

n=50, rank(L0)/n=0.2

(d) RMC1

0.1 0.2 0.3 0.4γ

0.7

0.8

0.9

ρ

n=50, rank(L0)/n=0.2

(e) RMC2

0.1 0.2 0.3 0.4γ

0.7

0.8

0.9

ρ

n=50, rank(L0)/n=0.2

(f) QMC

Figure 3: Recovery for varying sparsity γ (x-axis) and observed rate ρ (y-axis).

is the percentage of observed entries.Similar to synthetic data case, we compare our proposed quaternion matrix completion

method (QMC) with the other two real matrix completion methods. The first comparison(RMC1) is to apply the real matrix completion method to the unfolding color image ofsize n1-by-3n2 and the second comparison (RMC2) is to apply the real matrix completionmethod to each color channel and then combine the results together. In QMC, we set thevalue of λ = 1√

ρn(1)given in Theorem 3.1. In RMC1 and RMC2, we set their corresponding

values according to their exact recovery results in [3], i.e., λ = 1√3ρn(1)

and λ = 1√ρn(1)

respectively. The stopping criteria of the alternating direction method of multipliers arethat the norm of the successive iterates is less than 1 × 10−8 and the maximum number ofiteration is 5000.

In Figures 6, 8, 10 and 12, we show the color images completion results for ρ = 0.2and γ = 0.0, 0.1, 0.2. Both Peak Signal-to-Noise Ratio (PSNR) and Structural Similarityindex (SSIM) are also listed for the comparison of different methods. We find that theperformance of PSNR and SSIM by the proposed QMC algorithm is better than those byRMC1 and RMC2. Moreover, we see from the figures that the color images restored byQMC are visually better than those restored by RMC1 and RMC2. In Figures 7, 9, 11 and13 corresponding to some zoomed regions in Figures 6, 8, 10 and 12 respectively, both thecolor and the details restored by QMC are visually better than those by RMC1 and RMC2.For instance, the proposed QMC algorithm can provide more clear and detailed restoration

11

n=50, ρ=0.90

0.1 0.2 0.3rank(L

0)/n

0.6

0.5

0.4

0.3

0.2

0.1

γ

(a) RMC1

n=50, ρ=0.90

0.1 0.2 0.3rank(L

0)/n

0.6

0.5

0.4

0.3

0.2

0.1

γ

(b) RMC2

n=50, ρ=0.90

0.1 0.2 0.3rank(L

0)/n

0.6

0.5

0.4

0.3

0.2

0.1

γ

(c) QMC

n=50, ρ=0.80

0.1 0.2 0.3rank(L

0)/n

0.6

0.5

0.4

0.3

0.2

0.1

γ

(d) RMC1

n=50, ρ=0.80

0.1 0.2 0.3rank(L

0)/n

0.6

0.5

0.4

0.3

0.2

0.1

γ

(e) RMC2

n=50, ρ=0.80

0.1 0.2 0.3rank(L

0)/n

0.6

0.5

0.4

0.3

0.2

0.1

γ

(f) QMC

Figure 4: Recovery for varying rank rank(L0) (x-axis) and sparsity γ (y-axis).

in the green stripes in Figure 7; in the boundaries of green and red peppers in Figure 9; inthe shapes of red flower (the upper part) and in the small white regions (the lower part) inFigure 11; and in the yellow stigma of flower and the shape of red flower (the upper part)and in the boundary of orange flower (the lower part) in Figure 13.

5 Conclusion

As a summary, we have studied and analyzed robust quaternion matrix completion problem.In particular, we have presented a rigorous analysis for provable estimation of quaternionmatrix from a random subset of its corrupted entries. Under this quaternion framework,robust color images completion can be solved by a convex optimization problem, whichminimizes a nuclear norm of quaternion matrix which is a convex surrogate for a quaternionmatrix rank, and the l1-norm of sparse quaternion matrix entries. Numerical results havebeen demonstrated our theoretical results and shown that the performance of the proposedmethod for color images completion is better than the testing methods using the unfoldingand performing image completion separately in red, green and blue channels.

12

n=50, γ=0.10

0.1 0.2 0.3rank(L

0)/n

0.7

0.8

0.9

ρ

(a) RMC1

n=50, γ=0.10

0.1 0.2 0.3rank(L

0)/n

0.7

0.8

0.9

ρ

(b) RMC2

n=50, γ=0.10

0.1 0.2 0.3rank(L

0)/n

0.7

0.8

0.9

ρ

(c) QMC

n=50, γ=0.20

0.1 0.2 0.3rank(L

0)/n

0.7

0.8

0.9

ρ

(d) RMC1

n=50, γ=0.20

0.1 0.2 0.3rank(L

0)/n

0.7

0.8

0.9

ρ

(e) RMC2

n=50, γ=0.20

0.1 0.2 0.3rank(L

0)/n

0.7

0.8

0.9

ρ

(f) QMC

Figure 5: Recovery for varying rank rank(L0) (x-axis) and and observed rate ρ (y-axis).

References

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6 Appendix

6.1 Proof of Convex Envelope

Proposition 6.1. Suppose that A and B are two n× n quaternion matrices with singularvalues

ρ1 ≥ ρ2 ≥ · · · ≥ ρn, σ1 ≥ σ2 ≥ · · · ≥ σn,

respectively, then

Re (Tr (AB)) ≤n∑

r=1

ρrσr.

15

Proof: It follows from Theorem 2.1 that there exist two pairs of unitary matrices Ui andVi (i = 1, 2) such that

A = U1Σ1V∗1 , B = U2Σ2V

∗2 ,

where Σ1 = diag (ρ1, · · · , ρn) and Σ2 = diag (σ1, · · · , σn). For quaternion matrices A

and B, Tr (AB) = Tr (BA) is not valid in general. However, we can get Re (Tr (AB)) =Re (Tr (BA)). Here Re(·) is the real part of a matrix. Based on this property, we have

Re (Tr (AB)) = Re (Tr (U1Σ1V∗1U2Σ2V

∗2))

= Re (Tr (V∗2U1Σ1V

∗1U2Σ2))

= Re (Tr (UΣ1VΣ2)) = Re

(

n∑

r,s=1

ur,svs,rρsσr

)

,

where U = (ur,s)n×n= V∗

2U1 and V = (vr,s)n×n= V∗

1U2 are two unitary matrices. Hence

Re

(

n∑

r,s=1

ur,svs,rρsσr

)

≤∣

∣

∣

∣

∣

Re

(

n∑

r,s=1

ur,svs,rρsσr

)∣

∣

∣

∣

∣

≤n∑

r,s=1

|ur,s| |vs,r| ρsσr

≤ 1

2

n∑

r,s=1

|ur,s|2ρsσr +1

2

n∑

r,s=1

|vs,r|2ρsσr ≤n∑

r=1

ρrσr.

Below is the result which gives the convex envelope of the rank function over the set ofquaternion matrices with bounded norm.

Proposition 6.2. Denote S = X ∈ Qn1×n2 | ‖X‖ ≤ 1, then the convex envelope of thefunction φ (X) = rank (X) on S can be expressed as

φenvo (X) = ‖X‖∗ =

minn1,n2∑

i=1

σi (X) .

Proof: We prove this theorem by applying conjugate function. The conjugate of the quater-nion matrix rank function φ, with spectral norm less than or equal to 1 is defined as

φ∗ (Y) = sup‖X‖≤1

(Re (Tr (Y∗X))− rank (X)) .

Denote n(2) = minn1, n2, then it follows from Proposition 6.1 that

Re (Tr (Y∗X)) ≤n(2)∑

i=1

σi (Y) σi (X) , (12)

where σi (X) denotes the ith largest singular value of X. For any given quaternion matrixY, there exist unitary matrices UY and VY such that Y = UYΣY V

∗Y. Choose X =

UYΣXV∗Y, then

Re (Tr (Y∗X)) = Re (Tr (VYΣY U∗YUYΣXV∗

Y))

= Re (Tr (VYΣY ΣXV∗Y))

= Tr (ΣY ΣX)

=

n(2)∑

i=1

σi (Y)σi (X) ,

16

which indicates that the equality in (12) can be achieved. Then the conjugate of quaternionmatrix rank function can be rewritten as

φ∗ (Y) = sup‖X‖≤1

(n(2)∑

i=1

σi (Y) σi (X)− rank (X)

)

.

Now we first consider separated cases with the rank of X is from 0 to n2. If rank(X) = 0,then φ∗ (Y) = 0 for all Y. If rank (X) = r for 1 ≤ r ≤ n(2), then

φ∗ (Y) =

r∑

i=1

σi (Y)− r.

Combing these results together, we can get

φ∗ (Y) = max

0, σ1 (Y)− 1, · · · ,r∑

i=1

σi (Y)− r, · · · ,n(2)∑

i=1

σi (Y)− n(2)

.

Further, note that if σi(Y)− 1 > 0, then the sum is bigger, thus

φ∗ (Y) =

0,∑r

i=1 σi (Y)− r,‖Y‖ ≤ 1,

σr (Y) > 1 and σr+1 (Y) < 1.

The conjugate of φ∗ (Y) is defined as

φ∗∗ (Z) = supY

(Re(Tr (Z∗Y))− φ∗ (Y))

for all Z ∈ Qn1×n2 . By similar analysis, suppose that Z = UZΣZV∗Z and choose Y =

UZΣYV∗Z, then

φ∗∗ (Z) = supY

(n(2)∑

i=1

σi (Z)σi (Y)− φ∗ (Y)

)

.

We can consider two cases, ‖Z‖ ≥ 1 and ‖Z‖ ≤ 1. For the first case, ‖Z‖ ≥ 1, σ1 (Y) canbe choosen large enough so that φ∗∗ (Z) → ∞. Then we have

φ∗∗ (Z) = supY

(n(2)∑

i=1

σi (Z) σi (Y)− φ∗ (Y)

)

= supY

(n(2)∑

i=1

σi (Z) σi (Y)−(

r∑

i=1

σi (Y)− r

))

= supY

(

σ1 (Y) (σ1 (Z)− 1) +

(n(2)∑

i=2

σi (Z) σi (Y)−(

r∑

i=2

σi (Y)− r

)))

.

In the second case ‖Z‖ ≤ 1, if ‖Y‖ ≤ 1, then φ∗ (Y) = 0 and the supremum is achieved atσi (Y) = 1, i = 1, ..., n(2), yielding

φ∗∗ (Z) = sup

n(2)∑

i=1

σi (Z) = ‖Z‖∗ .

17

If ‖Y‖ > 1, then the argument of the sup is always smaller than the value given above. Infact,

n(2)∑

i=1

σi (Z) σi (Y)−r∑

i=1

(σi (Y) − 1)

=

n(2)∑

i=1


i=1

(σi (Y) − 1)−n(2)∑

i=1

σi (Z) +

n(2)∑

i=1

σi (Z)

=

r∑

i=1

σi (Z) σi (Y) +

n(2)∑

i=r+1


i=1

(σi (Y)− 1)−n(2)∑

i=1

σi (Z) +

n(2)∑

i=1

σi (Z)

=

r∑

i=1

(σi (Z) − 1) (σi (Y)− 1) +

n(2)∑

i=r+1

σi (Z) (σi (Y) − 1) +

n(2)∑

i=1

σi (Z)

≤n(2)∑

i=1

σi (Z) ,

which can be derived from the following two inequalities

r∑

i=1

(σi (Z)− 1) (σi (Y) − 1) ≤ 0 and

n(2)∑

i=r+1

σi (Z) (σi (Y) − 1) ≤ 0.

In summary, we can get

φ∗∗ (Z) = supY

n(2)∑

i=1

σi (Z) = ‖Z‖∗

over the set ‖Z‖ ≤ 1.

6.2 Proof of Lemma 3.2

In order to show Lemma 3.2, we need the following result which was originally proven byLust-Picquard [21], and was later sharpened by Buchholz [2].

Proposition 6.3. Let Xi1≤i≤r be a finite sequence of matrices of the same dimensionand εi be a Rademacher sequence. For each q ≥ 2

Eε

∥

∥

∥

∥

∥

∑

i

εiXi

∥

∥

∥

∥

∥

q

Sq

1q

≤ Ck

√qmax

∥

∥

∥

∥

∥

∥

(

∑

i

X∗i Xi

)12

∥

∥

∥

∥

∥

∥

Sq

,

∥

∥

∥

∥

∥

∥

(

∑

i

XiX∗i

)12

∥

∥

∥

∥

∥

∥

Sq

,

where Ck√q = 2−

14

√

π/e.

Proof of Lemma 3.2: Suppose that 1 ≤ q ≤ ∞. The Schatten-q-norm of an n-by-nquaternion matrix can be defined as

‖X‖Sq=

(

n∑

i=1

σi(X)q

)1q

,

18

where σi(X) are the singular values of the quaternion matrix X. For q ≥ logn, then byProposition 6.3, we have

‖X‖Sq‖ ≤ ‖X‖ ≤ e‖X‖Sq

.

Let ε1, ..., εN be independent Bernoulli variables taking values 1,−1 with probability 1/2and let x1, ..., xN , x′

1, ..., x′N be independent copies of x. Since xix

∗i − x′

ix′∗i is a Hermitian

random matrix,

Ex

∥

∥

∥

∥

∥

1

N

N∑

i=1

xix∗i − I

∥

∥

∥

∥

∥

≤ ExEx′

∥

∥

∥

∥

∥

1

N

N∑

i=1

xix∗i −

1

N

N∑

i=1

x′ix

′∗i

∥

∥

∥

∥

∥

≤ EεExEx′

∥

∥

∥

∥

∥

1

N

N∑

i=1

εi (xix∗i − x′

ix′∗i )

∥

∥

∥

∥

∥

≤ 2ExEε

∥

∥

∥

∥

∥

1

N

N∑

i=1

εixix∗i

∥

∥

∥

∥

∥

.

Applying Proposition 6.3, we can get

E

∥

∥

∥

∥

∥

N∑

i=1

εixix∗i

∥

∥

∥

∥

∥

≤ e

E

∥

∥

∥

∥

∥

N∑

i=1

εixix∗i

∥

∥

∥

∥

∥

p

Snq

1q

≤ eBq

∥

∥

∥

∥

∥

∥

(

N∑

i=1

‖xi‖2xix∗i

)

12

∥

∥

∥

∥

∥

∥

Snq

≤ eBq

∥

∥

∥

∥

∥

∥

(

N∑

i=1

‖xi‖2xix∗i

)

12

∥

∥

∥

∥

∥

∥

≤ eBq maxi=1,...,N

‖xi‖∥

∥

∥

∥

∥

N∑

i=1

xix∗i

∥

∥

∥

∥

∥

12

.

where Bq ≤ C√q is shown in [21] and [2]. The results follow.


Proof: If A = 0, then the result is obvious. We now assume A 6= 0 and rank(A) = r. LetY be given as (6). Then we have

Re (〈A,Y〉) = Re (Tr (A∗Y)) = Re (Tr (A∗ (direc(A) + F))) =∑

1≤k≤r

|ai,j | = ‖A‖1 ,

and‖Y‖∞ = ‖direc(A) + F‖∞ ≤ ‖direc(A)‖∞ , ‖F‖∞ = 1.

It follows form that Y is a subgrandient of the ‖·‖1 at A. On the other hand, Y ∈ ∂ ‖A‖1if and only if Y ∈ V (A : ‖·‖∞), where

V (A : ‖ · ‖∞) = G : G ∈ Qn1×n2 , Re〈A,G〉 = ‖A‖1, ‖G‖∞ = 1.

Suppose that A ∈ Qn1×n2 has the SVD as (1), then we can get

V (A : ‖ · ‖∞) = G : G = UDV∗, D ∈ V (Σ : ‖ · ‖∞).

Since Σ is a real matrix, D ∈ V (Σ : ‖·‖∞) if and only if D ∈ ∂ ‖Σ‖1, which can be expressedas follows:

D = sgn(U∗AV) + F1,

where PΩ(U∗AV) (F1) = 0 (the projection of the entries of F1 on the support of of U∗AV)and ‖F1‖∞ ≤ 1. The results follow.

19


Proof: If A = 0, then the result is obvious. We now assume A 6= 0. Let Y be given as (7)satisfying (a) and (b) in Lemma 3.4. It follows from that

Re (〈A,Y〉) = Re (Tr (A∗Y)) = Re

Tr

A∗

∑

1≤k≤r

ukv∗k + F

= Re

Tr

VΣU∗ ∑

1≤k≤r

ukv∗k

=∑

1≤k≤r

σi = ‖A‖∗ .

Moreover,

‖Y‖ =

∥

∥

∥

∥

∥

∥

∑

1≤k≤r

ukv∗k + F

∥

∥

∥

∥

∥

∥

= max

σ

∑

1≤k≤r

ukv∗k

, σ (F)

= 1.

Then Y is a subgrandient of the nuclear norm at A. On the other side, Y ∈ ∂‖A‖∗ if andonly if Y ∈ V (A : ‖ · ‖), where

V (A : ‖ · ‖) =

G : G ∈ Qn1×n2 , Re (〈A,G〉) = ‖A‖∗, ‖G‖ = 1

.

The SVD of A ∈ Qn1×n2 can be expressed as (1), thus

V (A : ‖ · ‖) = G : G = UDV∗, D ∈ V (Σ : ‖ · ‖) .

Since Σ is a real matrix, D ∈ V (Σ : ‖ · ‖) if and only if D ∈ ∂‖Σ‖∗, which can be expressedas

D = U1V∗1 + F1,

where the column spaces of W1 is orthogonal to U1 = spanu11, u12, · · · , u1r, and the rowspace of F1 is orthogonal to V1 = spanv11, v12, · · · , v1r; and the spectral norm of W1 isless than or equal to 1. Note that ui and vi i = 1, ..., r are parts of the standard bases ofthe vector space, then we can get Y ∈ ∂‖A‖∗ can be expressed as (7) satisfying (a) and (b)in Lemma 3.4. The result follows.


In the proof, we do not derive the results directly by the usual inner product of two quater-nion matrices. Indeed, we introduce an equivalent form of the inner product of quaternionmatrices as

〈A,B〉 =n1∑

i

n2∑

j

eTi a∗i,jBej,

where A,B ∈ Qn1×n2 , and ei is the i-th column of the identity matrix.Proof: By the definition of PT, we can get PT (A) = PUA+APV−PUAPV and PT (B) =PUB + BPV − PUBPV. Take them into 〈A,PT (B)〉 and choose ei as the i-th column of

20

the identity matrix, we have

〈A,PT (B)〉 = 〈A∗,PUB+BPV − PUBPV〉 =⟨

n1∑

i=1

n2∑

j=1

ai,jeieTj ,PUB+BPV − PUBPV

⟩

=

n1∑

i=1

n2∑

j=1

(⟨

ai,jeieTj ,PUB

⟩

+⟨

ai,jeieTj ,BPV

⟩

−⟨

ai,jeieTj ,PUBPV

⟩)

=

n1∑

i=1

n2∑

j=1

(

eTi a∗i,jPUBej + eTi a

∗i,jBPVej − eTi a

∗i,jPUBPVej

)

.

Similarly,

〈PT (A) ,B〉 = 〈PUA+APV − PUAPV,B〉

=

n1∑

i=1

n2∑

j=1

⟨

PUai,jeieTj + ai,jeie

Tj PV − PUai,jeie

Tj PV,B

⟩

=

n1∑

i=1

n2∑

j=1

(

eTi a∗i,jPUBej + eTi a

∗i,jBPVej − eTi a

∗i,jPUBPVej

)

.

The results follow.

6.6 Proof of Lemmas 3.6, 3.7, 3.8, 3.9 and 3.10

Firstly, we show the following proposition which will be used later.

Proposition 6.4. For each pair of quaternion matrices W and H,

Re (〈W,H〉) ≤ ‖W‖ ‖H‖∗ .

In addition, for any quaternion matrix H, there is a matrix W obeying ‖W‖ = 1, whichachieves the above equality.

Proof: Suppose that H has the SVD as H = UΣV∗, then

Re (〈W,H〉) = Re (〈W,UΣV∗〉) = Re (Tr (W∗UΣV∗)) = Re (Tr (V∗W∗UΣ)) ≤ ‖W‖ ‖H‖∗ .

If we take W = UV∗, then Re 〈W,H〉 = ‖W‖ ‖H‖∗ .

Proof of Lemma 3.6: The proof can be divided into two parts. (a) We first prove thatthere exists a constant c′r such that

E

(

ρ−1 ‖PTPΩPT − ρPT‖)

≤ c′r

√

µ0n(1)r logn(1)

m, (13)

provided that the last expression is smaller than 1. (b) Once (a) is satisfied, for each λ > 0,we have

Prob

(

|Z − E (Z)| > λ

√

µ0n(1)r logn(1)

m

)

≤ 3 exp

(

−τ ′0 min

λ2 logn(1), λ

√

m logn(1)

µ0n(2)r

)

,

(14)

21

where

Z ≡ ρ−1 ‖PTPΩPT − ρPT‖ = ρ−1∥

∥Σij (δij − ρ)PT

(

eieTj

)

⊗ PT

(

eieTj

)∥

∥ .

For (a), suppose that ei is the i-th column of the identity matrix, then

PT

(

eieTj

)

= PUeieTj + ei (PVej)

∗ − PUeieTj PV.

Recalling Lemma 3.5, we have

∥

∥PT

(

eieTj

)∥

∥

2

F=⟨

PT

(

eieTj

)

,PT

(

eieTj

)⟩

=⟨

PT

(

eieTj

)

, eieTj

⟩

=⟨

PUeieTj + ei (PVej)

∗ − PUeieTj PV, eie

Tj

⟩

=⟨

PUeieTj , eie

Tj

⟩

+⟨

ei (PVej)∗, eie

Tj

⟩

−⟨

PUeieTj PV, eie

Tj

⟩

= Tr(

ejeTi PUeie

Tj

)

+Tr(

PVejeTi eie

Tj

)

− Tr(

PVejeTi PUeie

Tj

)

= Tr((

eTi PUei)

ejeTj

)

+Tr(

PVeieTj

(

eTi ei))

− Tr(

PVejeTj

(

eTi PUei))

= ‖PUei‖2F + ‖PVej‖2F − ‖PUei‖2F ‖PVej‖2F .

Using both ‖PUei‖2F ≤ µ(U)rn1

and ‖PVej‖2F ≤ µ(V)rn2

, we can derive

∥

∥PT

(

eieTj

)∥

∥

2

F≤ 2µ0r

minn1, n2=

2µ0r

n(2).

Next, we will show (13) is satisfied. Note that any quaternion matrix X can be expandedas X =

∑

i,j

⟨

eieTj , X

⟩

eieTj , where (eieTj )1≤i,j≤n is the canonical basis. Therefore,

PT (X) =∑

i,j

⟨

eieTj ,PT (X)

⟩

eieTj and PΩPT (X) =

∑

i,j

δij⟨

eieTj ,PT (X)

⟩

eieTj .

It follows that

PTPΩPT (X) =∑

i,j

⟨

eieTj ,PT

∑

i,j

δi,j⟨

eieTj ,PT (X)

⟩

eieTj

⟩

eieTj

=∑

i,j

⟨

PT

(

eieTj

)

,∑

i,j

δij⟨

eieTj ,PT (X)

⟩

eieTj

⟩

eieTj

=∑

i,j

δi,j∑

i,j

⟨

PT

(

eieTj

)

,⟨

eieTj ,PT (X)

⟩

eieTj

⟩

eieTj

=∑

i,j

δi,j∑

i,j

⟨

PT

(

eieTj

)

, eieTj

⟩

eieTj

⟨

eieTj ,PT (X)

⟩

=∑

i,j

δi,jPT

(

eieTj

) ⟨

eieTj ,PT (X)

⟩

=∑

i,j

δi,jPT

(

eieTj

) ⟨

PT

(

eieTj

)

,X⟩

.

In other words,

PTPΩPT =∑

i,j

δi,jPT

(

eieTj

)

⊗ PT

(

eieTj

)

.

22

Thus,

PTPΩPT

(

eieTj

)

− ρPT

(

eieTj

)

=∑

i,j

(δi,j − ρ)PT

(

eieTj

)

⊗ PT

(

eieTj

)

.

Note that

E

∑

i,j

(

δi,j −m

n1n2

)

PT

(

eieTj

)

⊗ PT

(

eieTj

)

= 0,

and by Lemma 3.2, we have

E (Z) = E

ρ−1

∥

∥

∥

∥

∥

∥

∑

i,j

(δi,j − ρ)PT

(

eieTj

)

⊗ PT

(

eieTj

)

∥

∥

∥

∥

∥

∥

= ρ−1E

∥

∥

∥

∥

∥

∥

∑

ij

(δi,j − ρ)PT

(

eieTj

)

⊗ PT

(

eieTj

)

∥

∥

∥

∥

∥

∥

≤ c1n1n2

m

√m√logn1n2√n1n2

√

2µ0r

n(2)≤ c′r

√

µ0n(1)r logn(1)

m,

where c1 is defined as in Lemma 3.2 and c′r = 2c1.For (b), denote

Yi,j = ρ−1

(

δi,j −m

n1n2

)

PT

(

eieTj

)

⊗ PT

(

eieTj

)

,

and set the (i, j)-th entry of Y to be Σi,jYi,j . Then

Z = sup |〈X1,Y (X2)〉|

= sup ρ−1∑

i,j

∣

∣

∣

∣

⟨

X1,

(

δi,j −m

n1n2

)

PT

(

eieTj

)

⊗ PT

(

eieTj

)

(X2)

⟩∣

∣

∣

∣

= sup ρ−1∑

i,j

(

δi,j −m

n1n2

)

∣

∣

⟨

PT

(

eieTj

)

(X1) ,PT

(

eieTj

)

(X2)⟩∣

∣ ,

where the supremum is over a countable collection of matricesX1 andX2 obeying ‖X1‖F ≤ 1and ‖X2‖F ≤ 1. Moreover, we can get

|〈X1,Y (X2)〉| = ρ−1

∣

∣

∣

∣

∣

∣

∑

i,j

(

δi,j −m

n1n2

)

∣

∣

∣

∣

∣

∣

∣

∣

⟨

PT

(

eieTj

)

(X1) ,PT

(

eieTj

)

(X2)⟩∣

∣

≤ ρ−1

∣

∣

∣

∣

∣

∣

∑

i,j

(

δi,j −m

n1n2

)

∣

∣

∣

∣

∣

∣

∥

∥PT

(

eieTj

)∥

∥

2

F= ρ−1

∥

∥PT

(

eieTj

)∥

∥

2

F

≤ n1n2

m

2µ0r

n(2)≤ 2n(1)µ0r

m

23

and

E |〈X1,Y (X2)〉|2 = ρ−1 (1− ρ)∣

∣

⟨

PT

(

eieTj

)

(X1) ,PT

(

eieTj

)

(X2)⟩∣

∣

≤ ρ−1 (1− ρ)∥

∥PT

(

eieTj

)∥

∥

2

F

∣

∣

⟨

PT

(

eieTj

)

,X2

⟩∣

∣

2

≤ n1n2

m

2µ0r

n(2)

∣

∣

⟨

PT

(

eieTj

)

,X2

⟩∣

∣

2 ≤ 2n(1)µ0r

m.

By the Talagrand’s concentration inequality, we have

Prob (|Z − E(Z)| > t) ≤ 3 exp

(

− t

KBlog

(

1 +t

2

))

≤ 3 exp

(

− t log 2

KBmin

1,t

2

)

,

where t = λ

√

µ0n(1)r logn(1)

mand B =

2n(1)µ0r

m. Set λ =

√

βθ

and assume that m >βτ ′0µ0n(1)r logn(1), then the right hand of (14) is upper bounded by 3n−β

(1) . Thus we canget

Z ≤ c′r

√

µ0n(1)r logn(1)

m+

√

β

θ

√

µ0n(1)r log n(1)

m,

which means that (9) is satisfied with c = c′r +√

βθ.

By using Lemma 3.6, we can prove Lemma 3.7, which plays an important role in thedual certification.

Proof of Lemma 3.7: Choosing

ε ≥ c

√

µ0n(1)r logn(1)

m,

and note that ρ = mn1n2

, we can get

ε2 ≥ c2µ0n(1)r logn(1)

m× n(2)

n(2)= c2

µ0ρ−1r logn(1)

n(2).

It follows form (9) that if ρ ≥ c2µ0r logn(1)

ε2n(2), then (10) is satisfied.

Proof of Lemma 3.8: From Lemma 3.7, we have∥

∥

∥PT − (1− ρ)−1 PTPΩ⊥PT

∥

∥

∥ ≤ ε

with

(1− ρ) ≥ c2µ0r logn(1)

ε2n(2).

Since I = PΩ + PΩ⊥ ,

PT − (1− ρ)−1 PTPΩ⊥PT = (1− ρ)

−1((1− ρ)PT − PTPΩ⊥PT) = (1− ρ)

−1(PTPΩPT − ρPT) .

Thus(PTPΩPT − ρPT) = (1− ρ)

(

PT − (1− ρ)−1 PTPΩ⊥PT

)

andPTPΩPT = ρPT + (1− ρ)

(


)

.

24

It follows from that

‖PTPΩPT‖ ≤ ‖ρPT‖+∥

∥

∥(1− ρ)(


)∥

∥

∥ ≤ ρ+ (1− ρ) ε ≤ ρ+ ε.

Moreover, when ρ+ ε ≤ 14 , we can get ‖PΩPT‖ ≤ 1

2 .

Proof of Lemma 3.9: For two arbitrary quaternion matrices A and B with proper sizes,

‖AB‖F ≤ ‖A‖ ‖B‖F .

Then for any Z ∈ T, we can get Z = PT (Z) , and∥


∥

F=∥

∥PT (Z)− ρ−1PTPΩPT (Z)∥

∥

F=∥

∥

(

PT − ρ−1PTPΩPT

)

PT (Z)∥

∥

F

≤∥

∥

(

PT − ρ−1PTPΩPT

)∥

∥ ‖PT (Z)‖F ≤ ε ‖Z‖F .

Proof of Lemma 3.10: Define Ω = (i, j) : δi,j = 1 where δi,j is sampled from theBernoulli model with parameter ρ. Denote Z′ = Z−ρ−1PTPΩZ, which also can be given by

Z′ =∑

i,j

(1 − ρ−1δi,j)PT

(

ZijeieTj

)

,

where ei, ej, i, j = 1, ..., n are defined as in Lemma 3.6. Then Z′i0,j0

can be expressed as asum of independent random variables, i.e.

Z′i0j0

= ΣijYij , Yi,j =∑

i,j

(1 − ρ−1δi,j)⟨

PT

(

ZijeieTj

)

, ei0eTj0

⟩

.

It is easy to prove E (Yi,j) = 0, and

∑

i,j

V ar (Yi,j) =∑

i,j

E

(

(1− ρ−1δi,j)⟨

PT

(

ZijeieTj

)

, ei0eTj0

⟩)2.

Note that

E

(

(1− ρ−1δi,j)2)

= E

(

1− 2ρ−1δij + ρ−2δ2i,j)

=E

(

ρ2 − 2ρδi,j + δ2i,j)

ρ2= (1− ρ) ρ−1.

From the proof of Lemma 3.6, we have

|Yi,j | = |∑

i,j

(1− ρ−1δi,j)⟨

PT

(

ZijeieTj

)

, ei0eTj0

⟩

| ≤ 2ρ−1‖Z‖∞µ0r/n(2),

and

∑

i,j

V ar (Yi,j) ≤ (1− ρ) ρ−1

∑

i,j

∣

∣

⟨

PT

(

ZijeieTj

)

, ei0eTj0

⟩∣

∣

2

= (1− ρ) ρ−1(

Σij

∣

∣

⟨

ZijeieTj , PT

(

ei0eTj0

)⟩∣

∣

2)

≤ (1− ρ) ρ−1(

‖Z‖2∞∥

∥PT

(

ei0eTj0

)∥

∥

2

F

)

≤ (1− ρ) ρ−1‖Z‖2∞2µ0r

n(2).

Then by the Bernstein inequality, we have

Prob (|Zij | > ε ‖Z‖∞) ≤ 2 exp

(

− ε2n(1)ρ

4 (1− ρ+ ε)µ0r

)

.

It follows form that (11) is satified when ρ satisfies the assumption.

25

6.7 Proof of Theorem 3.11

Proof of Theorem 3.11: First, we can get Γ⊥∩T = 0 by ‖PTPΓ⊥M‖F ≤ n(1) ‖PT⊥PΓ⊥M‖F .Next we consider a feasible perturbation (L0 +HL,S

′0 −HS) where PΩHL = PΩHS , and

show that the objective increases unlessHL = HS = 0. In other words, (L0,S′0) is the unique

solution. Let UV∗ +W0 be a subgradient of the nuclear norm at L0, and direc(S′0) + F0

be a subgradient of the l1 norm at S′0. It follows from Lemmas 3.3-3.4 that

‖L0 +HL‖∗ + λ ‖S′0 −HS‖1 ≥ ‖L0‖∗ + λ ‖S′

0‖1 + Re (〈UV∗ +W0,HL〉)−λRe (〈direc (S′

0) + F0,HS〉) .

Choose W0 = PT⊥(U1V∗1), where PT⊥HL = U1ΣV

∗1 , and F0 = −direc(PΓHS), we can

get Re (〈W0,HL〉) = ‖PT⊥HL‖∗ and Re (〈F0,HS〉) = −‖PΓHS‖1. Then

‖L0 +HL‖∗+λ ‖S′0 −HS‖1 ≥ ‖L0‖∗+λ ‖S′

0‖1+1

2(‖PT⊥HL‖∗ + λ ‖PΓHL‖1)−

1

n2(1)

‖PTHL‖F .

Note that

‖PTHL‖F ≤ ‖PTPΓHL‖F + ‖PTPΓ⊥HL‖F ≤ ‖PTPΓHL‖F + n(1) ‖PT⊥PΓ⊥HL‖F≤ ‖PTPΓHL‖F + n(1) ‖PT⊥ (I − PΓ)HL‖F≤ ‖PTPΓHL‖F + n(1) (‖PT⊥HL‖F + ‖PT⊥PΓHL‖F )≤ ‖PTPΓHL‖F + n(1) ‖PT⊥PΓHL‖F + n(1) ‖PT⊥HL‖F≤ (n(1) + 1) ‖PΓHL‖F + n(1) ‖PT⊥HL‖F .

Applying ‖PΓHL‖F ≤ ‖PΓHL‖1 and ‖PT⊥HL‖F ≤ ‖PT⊥HL‖∗, we can get

‖L0 +HL‖∗+λ ‖S′0 −HS‖1 ≥ ‖L0‖∗+λ ‖S′

0‖1+(

1

2− 1

n(1)

)

‖PT⊥HL‖∗+(

λ

2− n(1) + 1

n2(1)

)

‖PΓHL‖1 .

Then the claim follows form Γ⊥ ∩ T = 0.

By using the results in Lemmas 3.6–3.10, we show high probability that the assumptionin Theorem 3.11 is valid.

Proposition 6.5. Under the assumptions of Theorem 3.1, the assumption of Theorem 3.11is satisfied. That is, for any quaternion matrix M, ‖PTPΓ⊥M‖F ≤ n(1) ‖PT⊥PΓ⊥M‖F holdswith high probability.

Proof: Set ρ0 = ργ, then Γ is characterized by the Bernoulli probability distribution withparameter ρ0. From Lemma 3.7,

∥

∥PT − ρ−10 PTPΓPT

∥

∥ ≤ 12 with high probability. Further,

from‖PΓPTPΓ⊥M‖F = ‖PΓ (I − PT⊥)PΓ⊥M‖F = ‖PΓPT⊥PΓ⊥M‖F ,

we have‖PΓPTPΓ⊥M‖F ≤ ‖PT⊥PΓ⊥M‖F .

Moreover,〈PΓPTPΓ⊥M,PΓPTPΓ⊥M〉 = 〈PTPΓ⊥M,PTPΓPTPΓ⊥M〉,

26

implies that

ρ−10 ‖PΓPTPΓ⊥M‖2F = ρ−1

0 〈PΓPTPΓ⊥M,PΓPTPΓ⊥M〉=‖〈PTPΓ⊥M, ρ−1

0 PTPΓPTPΓ⊥M〉‖=‖〈PTPΓ⊥M,PTPΓ⊥M〉+ 〈PTPΓ⊥M,

(

ρ−10 PTPΓPT − PT

)

PΓ⊥M〉‖

≥ ‖PTPΓ⊥M‖2F − 1

2‖PTPΓ⊥M‖2F =

1

2‖PTPΓ⊥M‖2F .

Therefore, ‖PT⊥PΓ⊥M‖2F ≥ ‖PΓPTPΓ⊥M‖2F ≥ ρ0

2 ‖PTPΓ⊥M‖2F . Note that ρ0 = mγn1n2

, then√

ρ0

2 ≥√

mγ

2

n(1)≥ 1

n(1)is satisfied if mγ > 2. Thus, we can derive the claim.

6.8 Construction of YL and WS

The final step of the proof is to construct YL and WS .

6.8.1 Construction of YL via the Golfing Scheme

We use the similar golf scheme, which was introduced by Gross [13], and modified by Candes[4], to construct the dual certificate YL.

It follows from the sampling scheme that the available and clean data Γ is characterizedby the Bernoulli probability distribution with parameter ρ0 = γρ. Moreover Γ can also beexpressed as Γ = ∪1≤j≤j0Γj, where Γj , j = 1, ..., j0 are independent and can be characterizedby the Bernoulli probability distribution with parameter q. Note that q obeys ρ0 = 1− (1−q)j0 . Take j0 = [3 logn(1)], then q ≥ ρ0

j0. We can inductively define

Yj = Yj−1 + q−1PΓjPT (UV∗ −Yj−1) ,

with Y0 = 0, and setting YL = Yj0 .

6.8.2 Construction of WS via the Least Square method

Note that unreliable data Ω′ is characterized by the Bernoulli probability distribution withparameter (1 − γ)ρ, then it follows from Lemma 3.8 that ‖PΩ′PT‖ ≤

√

(1 − γ)ρ+ ε holdswith high probability, where ε is a function of γ. Recalling that Ω is characterized by theBernoulli probability distribution with parameter ρ, then by Lemma 3.6 we can derive therestriction of PTPΩPT to T is invertible with high probability. And ‖(PTPΩPT)

−1‖ ≤ 2ρ−1

follows form PTPΩPT ≥ ρ2PT directly. Because

PΩ′ (PT + PΩ⊥)PΩ′ = PΩ′PT(PTPΩPT)−1PTPΩ′ ,

thus ‖PΩ′ (PT + PΩ⊥)PΩ′‖ ≤ 2ρ−1(ρ(1− γ) + ε) with high probability. We then set

WS = λ(I − (PT + PΩ⊥))(PΩ′ − PΩ′ (PT + PΩ⊥)PΩ′)−1direc(S′0).

Next, we show YL and WS satisfy (a)-(i) in Section 3.2 in details, respectively.

(a) By the definition of YL, we can get YL ∈ Γ, then PΓ⊥YL = 0 is derived.

27

(b) Denote Zj = UV∗ − PT (Yj) , then

Zj = UV∗ − PT (Yj) = UV∗ − PT

(

Yj−1 + q−1PΓjPT (UV∗ −Yj−1)

)

= UV∗ − PT (Yj−1)− q−1PTPΓjPT (UV∗ −Yj−1)

= PT (UV∗)− PT (Yj−1)− q−1PTPΓjPT (UV∗ −Yj−1)

= PT (UV∗ −Yj−1)− q−1PTPΓjPT (UV∗ −Yj−1)

=(

PT − q−1PTPΓjPT

)

(UV∗ −Yj−1)

=(

PT − q−1PTPΓjPT

)

Zj−1,

and YL = Yj0 = q−1∑

j PΓjZj−1. By Lemmas 3.9-3.10, if q ≥ c2

µ0r logn(1)

ε2n(2), then

‖Zj‖ ≤ ej ‖UV∗‖∞ , ‖Zj‖F ≤ ej ‖UV∗‖F = ej√r.

From ‖UV∗‖∞ ≤√µ0r

n(2)and j0 = [3 logn(1)], we can deduce

‖PT⊥Yj0‖ =

∥

∥

∥

∥

∥

∥

∑

j

q−1PT⊥PΓjPT (Zj−1)

∥

∥

∥

∥

∥

∥

≤∑

j

∥

∥q−1PT⊥PΓjPT (Zj−1)

∥

∥

≤∑

j

∥

∥q−1PΓj(Zj−1)− Zj−1

∥

∥ =∑

j

∥

∥q−1(

PΓj− I

)

(Zj−1)∥

∥

≤ c2

√

n(1) logn(1)

q

∑

j

‖Zj−1‖∞ ≤ c2

√

n(1) logn(1)

q

∑

j

εj ‖UV∗‖∞

= c2

√

n(1) logn(1)

q× ε

(

1− εj)

1− ε×

√µ0r

n(2)≤ c′

√

µ0r log2 n(1)

n(2)ρ0.

When ρ0 ≥ 16(c′)2µ0r log2 n(1)

n(2), then

∥

∥PT⊥Y L∥

∥ ≤ 14 can be derived (which is possible

provided that ρr in Theorem 3.1 is sufficiently small).

(c) By the definitions of YL and Zj0 , we can get

∥

∥PTYL −UV∗∥

∥

F= ‖Zj0‖F ≤ ej0

√r ≤ e−3 log n(1)

√r ≤ n−2

(1).

(d) Similarly,

∥

∥PΓYL∥

∥

∞ ≤∥

∥YL∥

∥

∞ ≤ q−1 ‖UV∗‖∞∑

j

e−j ≤ 3(

1− e−1)

√

√

√

√

µ0r log2 n(1)

n2(2)ρ

20

.

Then we can derive (d) by setting λ4 = 1

4

√

γn(2)ρ0

.

(e),(f),(g) Denote E = direc(S′0), which distributes as

Ei,j =

ai,j =(

(S′

0)i,j|(S′

0)i,j |

)

, with probability ρ(1− γ).

0, with probability1− ρ(1− γ).

28

Then we can construct WS as

WS = λ(I − (PT + PΩ⊥))(PΩ′ − PΩ′ (PT + PΩ⊥)PΩ′)−1E

= (I − (PT + PΩ⊥))(WS0 +WS

1 ),

where WS0 = λE and WS

1 = λΣk≥1 (PΩ′ (PT + PΩ⊥)PΩ′)kE. It is easy to prove

(e), (f) and (g) are satisfied by the definition of WS .

(h) Because the entries of E are i.i.d and take zero with probability 1 − ρ − ργ, by theresults in [33] we can get ‖E‖ ≤ 4

√

n(1)ρ(1− γ) with large probability. In other

words, when λ = 1√n(1)ρ

,∥

∥WS0

∥

∥ ≤ 4√1− γ < 1

8 holds if 1− τ is small enough. Denote

Φ =∑

k≥1 (PΩ′ (PT + PΩ⊥)PΩ′)k. Obviously, Φ is Hermitian. In order to consider the

norm of the operator, we need the standard covering theory. Known from Theorem4.16 in Ledoux [17], the size of 1/2-net denoted by N for Euclidean sphere Sn−1 is atmost 6n. Replacing the Euclidean sphere with the 1/2-net, we obtain

‖Φ (E)‖ = supx,y∈Sn−1

|〈y,Φ (E)x〉| ≤ 4 supx,y∈N

|〈y,Φ (E)x〉| .

DenoteX (x,y) : |〈y,Φ (E)x〉| ,

where (x,y) is a pair of fixed unit vectors in N ×N . Since

|〈y,Φ (E)x〉| = |y∗Φ (E)x| = |Tr (xy∗Φ (E))| = |〈Φ (E) ,xy∗〉| = |〈Φ (xy∗) ,E〉| ,

we haveX (x,y) = |〈Φ (xy∗) ,E〉| .

Denote Ω′ = supp(S′0), then by the Hoeffding’s inequality

Prob (|X (x,y)| > t |Ω′ ) ≤ 2 exp

(

− 2t2

‖Φ (xy∗)‖2F

)

.

It follows from ‖xy∗‖F = 1 that ‖Φ (xy∗)‖F ≤ ‖Φ‖ . Then

Prob (‖Φ (E)‖ > t |Ω′ ) = Prob

(

4 supx,y∈N

|〈y,Φ (E)x〉| > t |Ω′)

≤ 2 |N |2 exp(

− t2

8 ‖Φ‖2

)

.

Moreover, ‖PΩ′ (PT + PΩ⊥)PΩ′‖ ≤ 2ρ−1(ρ(1− γ) + ε) implies

‖PΩ′ (PT + PΩ⊥)‖ ≤√

2(1− γ) + 2ρ−1ε = σ,

and ‖Φ‖ ≤ Σk≥1σ2k = σ2

1−σ2 . Setting α = σ2

1−σ2 , then we have

Prob (λ ‖Φ (E)‖ > t) ≤ 2 |N |2 exp(

− t2

2α2λ2

)

+ Prob (‖PΩ′ (PT + PΩ⊥)‖ ≥ σ) .

Since ‖(I − (PT + PΩ⊥))WS1 )‖ ≤ λ‖Φ(E)‖, ‖(I − (PT + PΩ⊥))WS

1 )‖ ≤ λ < 18 can be

derived when 2ρ−1(ρ(1 − γ) + ε) which is equivalent to 1− γ sufficiently small. Thenwhen λ = 1/

√n(1)ρ, we can derive ‖WS‖ ≤ 1

4 , if setting 1− γ sufficiently small.

29

(i) LetX(i,j) = (PΩ′ − PΩ′ (PT + PΩ⊥)PΩ′)−1PΩ′(I − (PT + PΩ⊥))eie

Tj .

Then for (i, j) ∈ Γ, we have

WSij =

⟨

eieTj ,W

S⟩

= λ⟨

eieTj , (I − (PT + PΩ⊥)) (PΩ′ − PΩ′ (PT + PΩ⊥)PΩ′)

−1E⟩

= λ⟨

X(i,j),E⟩

.

SincePΩ′(I − (PT + PΩ⊥))eie

Tj = PΩ′PT(PTPΩPT)

−1PTeieTj ,

and∥

∥PT

(

eieTj

)∥

∥

2

F≤ 2µ0r

n(2),

‖PΩ′(I − (PT + PΩ⊥))eieTj ‖F ≤ 2

√

(1− γ)ρ+ ε

ρ‖PTeie

Tj ‖F ≤ 2

√

2µ0r((1 − γ)ρ+ ε)

ρ√n(2)

holds with high probability. Moreover, it is easy to prove

‖ (PΩ′ − PΩ′ (PT + PΩ⊥)PΩ′)−1 ‖ ≤ 1

1− η,

where η = 2ρ−1(ρ(1− γ) + ε). It follows from that

sup(i,j)∈Γ

‖X(i, j)‖F ≤ 2√

2µ0r((1 − γ)ρ+ ε)

ρ(1− η)√n(2)

.

Applying the Hoeffding’s inequality again, we can get

Prob

(

sup(i,j)∈Γ

|WSij | >

λ

4

)

≤ 2n1n2 exp

(

− 1

8σ2

)

+ Prob

(

sup(i,j)∈Γ

‖X(i, j)‖F > σ

)

.

This shows that ‖PΓWS‖∞ ≤ λ/4 holds if 1− γ is sufficiently small.

30

(a) Original

(b) Observed (c) RMC1(28.96,0.9814) (d) RMC2(29.26,0.9833) (e) QMC(30.20,0.9869)

(f) Observed (g) RMC1(29.05,0.9810) (h) RMC2(29.38,0.9832) (i) QMC(30.22,0.9865)

(j) Observed (k) RMC1(29.07,0.9792) (l) RMC2(29.41,0.9814) (m) QMC(30.17,0.9850)

Figure 6: Visual comparison of completion results by different methods for NGC6543aimage with ρ = 0.2 and γ = 0.0, 0.1, 0.2 (from top to bottom). Here the two numbers ineach bracket refer to the values of PSNR and SSIM respectively.

31

(a) Original

(b) Observed (c) RMC1 (d) RMC2 (e) QMC

(f) Observed (g) RMC1 (h) RMC2 (i) QMC

(j) Observed (k) RMC1 (l) RMC2 (m) QMC

(a) Original




Figure 7: Zoomed blue part (the upper 13 pictures) and pink part (the lower 13 pictures)of completion results in Figure 6(a)-(m) respectively.

32

(a) Original

(b) Observed (c) RMC1(26.62,0.9620) (d) RMC2(28.20,0.9726 (e) QMC(29.61,0.9807)



Figure 8: Visual comparison of completion results by different methods for Pepper imagewith ρ = 0.2 and γ = 0.0, 0.1, 0.2 (from top to bottom). Here the two numbers in eachbracket refer to the values of PSNR and SSIM respectively.

33

(a) Original




(a) Original





34

(a) Original

(b) Observed (c) RMC1(25.74,0.9717) (d) RPCA2(26.37,0.9764) (e) QRCIC(26.97,0.9805)



Figure 10: Visual comparison of completion results by different methods for BSR124084image with ρ = 0.2 and γ = 0.0, 0.1, 0.2 (from top to bottom). Here the two numbers ineach bracket refer to the values of PSNR and SSIM respectively.

35

(a) Original




(a) Original





36

(a) Original

(b) Observed (c) RMC1(23.02,0.8809) (d) RMC2(23.76,0.8984) (e) QMC(24.61,0.9219)



Figure 12: Visual comparison of completion results by different methods for Flower imagewith ρ = 0.2 and γ = 0.0, 0.1, 0.2 (from top to bottom). Here the two numbers in eachbracket refer to the values of PSNR and SSIM respectively.

37

(a) Original




(a) Original





38

Documents

AQuaternionFrameworkforRobustColorImages Completionmng/mng_files/QMC.pdfperforming image completion separately in each color channel. Keywords: Color images, quaternion, matrix recovery,